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Composition Of Functions & Inverse Of A Function

Composite Functions

When two functions are combined in such a way that the output of one function becomes the input to another function, then this is referred to as composite function.

Composite Function

Consider three sets X, Y and Z and let f : X → Y  and g: Y → Z.

According to this, under map f, an element  x ∈ X is mapped to an element y = f(x) ∈ Y which in turn is mapped by g to an element z ∈ Z  in such a manner that z = g(y) = g[(f(x)] .

This mapping comprising of mappings f and g is known as composition of mappings. It is denoted by gof . Therefore, we are mapping onto .

The composite function is denoted as:

\(~~~~~~~~~\) ( gof)(x) = g(f (X) )

Similarly, (fog) (x) = f (g(x))

So, to find (gof) (x), take f(x)  as argument for the function g.

Let us try to solve some questions based on composite functions.

Let’s Work Out:

Example: Given the function f(x) = 3x + 5  and g(x) = \( 2x^3 \)  .Find ( gof)(x) and ( fog)(x).

Solution:  We know, (gof) (x) = g(f(x)) = g (3x+5) = \( 2(3x+5)^3 \)

Using Binomial Expansion, we have

\((gof) (x) = 2 \left [ (3x)^{3} + 3.(3x)^{2}.5 + 3.(3x)(5)^{2}+(5)^3 \right ]\)

\(\Rightarrow (gof) (x) = 2 \left [ 27x^{3} + 135x^{2} + 225 x + 125 \right]\)

\(\Rightarrow (gof) (x) = 54x^{3} + 270x^{2} + 450 x + 250 \)

Now, (fog)(x) = f \(\left( g(x) \right)\) = f ( \(2x^3\)) = \(3 (2x^{3}) + 5 \)

\(\Rightarrow (fog)(x) = 6x^{3} + 5\)

Example: Let f(x) = \(x^2\) and g(x) = \( \sqrt{1 – x^2}\) .Find (gof)(x) and ( fog)(x) .

Solution: (gof)(x) = g( f(x)) = g(\(x^2\)) = \( \sqrt{1 – (x^2)^2}\) = \( \sqrt{1 – x^4} \)

(fog) (x) = f(g(x)) = f( \( \sqrt{1 – x^2} \)) = ( \( \sqrt{1 – x^2})^2\) = \( 1 – x^2 \)

Inverse of a Function

Let f:X → Y. Now, let f represent a one to one function and y be any element of Y , there exists a unique element x ∈ X  such that y = f(x).Then the map

\(~~~~~~~~~~\) \( f^{-1}:f[X] \rightarrow X \)

which associates to each element is called as the inverse map of f.

The function f(x) = \( x^5 \) and g(x) = \( x^{\frac 12} \)   have the following property:

\( f\left( g(x) \right) \) = \( f \left( x^{\frac 15} \right) \) = \( (x^{\frac 15} )^5 \) = x

\( g\left( f(x) \right) \) = \( g \left( x^{5} \right) \) = \( (x^{5} )^{\frac 15} \) = x

Thus, if two functions f and g  satisfy \( f \left( g(x) \right) \) = x for every x in domain of f , then in such a situation we can say that the function f is the inverse of g  and g  is the inverse of f .

For finding the inverse of a function,we write down the function y as a function of x  i.e. y = f(x)   and then solve for x  as a function of y.

To have a better insight on the topic let us go through some examples.

Let’s Work Out:

Example: If f(x) = \(x^2\), g(x) = \( \frac{x}{3} \)   and h(x) = 3x+2 . Find out fohog(x).

Solution:  \( h\left( g(x) \right) \) = 3 \( \left ( \frac x3 \right) \) + 2 = x + 2

fohog(x) = f \( \left[ h\left(g(x)\right)\right] \) = \( ( x+ 2)^2 \)

This is the required solution.

Example: Example 2: Find the inverse of the function f(x) = \( x^3 \) , x ∈ R.

Solution: The given function f(x) = \( x^3 \)   is a one to one and onto function defined in the range → R  . Therefore, we can find the inverse of this function.

To find the inverse, we need to write down this function as

\(~~~~~~~~~~~~~\) y = \( x^3 \)

In the above equation,y  is an arbitrary element from the range of f. If we solve for x from the above equation, we will get:

\(~~~~~~~~~~~~~\) x = \( y^{\frac 13} \)

This gives a function g:Y →X. This new function g can be defined as

\(~~~~~~~~~~~~~\) g(y) = \( y^{\frac 13} \)

This function g is the inverse of the function f since it’s domain is same as the range of the function f.Since g(y) = \( y ^{\frac 12} \) , representing the independent variable with x , we get g(x) = \( x^{\frac 13}\) = \( f^{-1}(x) \)..

So, wasn’t that easy and fun? We have learnt about composition of functions and inverse functions. To learn more about functions, please visit our website www.byjus.com or download our app-BYJU’s the learning app. Happy Learning!!

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Practise This Question

The relation R is defined on the set of natural numbers as {(a,b) : a = 2b}. Then R−1 is given by 

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